Search any question & find its solution
Question:
Answered & Verified by Expert
Three capacitors of capacitances \(C_1=2 \mu \mathrm{F}\), \(C_2=3 \mu \mathrm{F}\) and \(C_3=5 \mu \mathrm{F}\) are connected in series. A potential difference of \(155 \mathrm{~V}\) is applied across the combination. Choose the correct option.
Options:
Solution:
2522 Upvotes
Verified Answer
The correct answer is:
Least potential difference is across \(C_3\). Equivalent capacitance of combination is \(\left(\frac{30}{31}\right) \mu \mathrm{F}\). The voltage across \(C_1\) is \(75 \mathrm{~V}\).
The given situation is shown in the following figure.
Equivalent capacitance is given as
\(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30} \Rightarrow C=\frac{30}{31} \mu \mathrm{F}\)
In series combination of capacitances, charge flowing through each capacitor is same which is given by
\(\begin{aligned}
q & =C V \\
& =\frac{30}{31} \times 155=150 \mu \mathrm{C}=1.5 \times 10^{-4} \mathrm{C}
\end{aligned}\)
\(\therefore\) Potential difference across \(C_1\),
\(\begin{aligned}
V_1 & =\frac{q}{C_1}=\frac{1.5 \times 10^{-4}}{2 \times 10^{-6}}=0.75 \times 10^2 \\
\Rightarrow V_1 & =75 \mathrm{~V}
\end{aligned}\)
Potential difference across \(C_2\),
\(\begin{aligned}
V_2 & =\frac{q}{C_2}=\frac{1.5 \times 10^{-4}}{3 \times 10^{-6}}=0.5 \times 10^2 \\
\Rightarrow V_2 & =50 \mathrm{~V}
\end{aligned}\)
Potential difference across \(C_3\),
\(V_3=\frac{q}{C_3}=\frac{1.5 \times 10^{-4}}{5 \times 10^{-6}}=0.3 \times 10^2 \Rightarrow V_3=30 \mathrm{~V}\)
Hence, potential difference across \(C_3\) is least as \(30 \mathrm{~V}\).
Equivalent capacitance is given as
\(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30} \Rightarrow C=\frac{30}{31} \mu \mathrm{F}\)
In series combination of capacitances, charge flowing through each capacitor is same which is given by
\(\begin{aligned}
q & =C V \\
& =\frac{30}{31} \times 155=150 \mu \mathrm{C}=1.5 \times 10^{-4} \mathrm{C}
\end{aligned}\)
\(\therefore\) Potential difference across \(C_1\),
\(\begin{aligned}
V_1 & =\frac{q}{C_1}=\frac{1.5 \times 10^{-4}}{2 \times 10^{-6}}=0.75 \times 10^2 \\
\Rightarrow V_1 & =75 \mathrm{~V}
\end{aligned}\)
Potential difference across \(C_2\),
\(\begin{aligned}
V_2 & =\frac{q}{C_2}=\frac{1.5 \times 10^{-4}}{3 \times 10^{-6}}=0.5 \times 10^2 \\
\Rightarrow V_2 & =50 \mathrm{~V}
\end{aligned}\)
Potential difference across \(C_3\),
\(V_3=\frac{q}{C_3}=\frac{1.5 \times 10^{-4}}{5 \times 10^{-6}}=0.3 \times 10^2 \Rightarrow V_3=30 \mathrm{~V}\)
Hence, potential difference across \(C_3\) is least as \(30 \mathrm{~V}\).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.